Answer:
y-3
Problem:
What is the remainder when the dividend is xy-3, the divisor is y, and the quotient is x-1. ?
Step-by-step explanation:
Dividend=quotient×divisor+remainder
So we have
xy-3=(x-1)×(y)+remainder
xy-3=(xy-y)+remainder *distributive property
Now we just need to figure out what polynomial goes in for the remainder so this will be a true identity.
We need to get rid of minus y so we need plus y in the remainder.
We also need minus 3 in the remainder.
So the remainder is y-3.
Let's try it out:
xy-3=(xy-y)+remainder
xy-3=(xy-y)+(y-3)
xy-3=xy-3 is what we wanted so we are done here.
Had -9-x^2
———
x^2
adh (-9-x^2)
——————-
x^2
-9adh-adhx^2
———————
x^2
9adh+adhx^2
- ———————
x^2
Hope this helps
Hi!
a: <em>F </em><em>n </em><em>G </em>=(Intersection) {4,6,8,10}
b: <em>F </em><em>n </em><em>H </em>={empty set}
Because there is no common element between the elements of set F and the elements of set H.
Note: It may not be very clear as I don't speak English.
God work!
